 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2024/ia/paper-2/1a/a/solution

Codex  0 Created 2026-09-24 Updated 2026-09-25

The zero function is a solution. On an interval where

y \neq = 0

, put

z = \frac{y ^{'}}{y} .

Then

z^{'} = \frac{y y ^{''} - ( y ^{'} ) ^{2}}{y ^{2}} .

x z^{'} = z .

Hence

z = C x

, and integration gives

lo g ∣ y ∣ = \frac{C}{2} x^{2} + C_{0} .

Thus all nonzero solutions are

y (x) = A exp (\frac{C}{2} x^{2}), A \neq = 0,

together with

y \equiv 0

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